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15 PROBLEM SOLUTIONS: Chapter Problem 2.1 At 60 Hz, ω = 120π primary: (Vrms )max = N1 ωAc (Brms )max = 2755 V, rms secondary: (Vrms )max = N2 ωAc (Brms )max = 172 V, rms At 50 Hz, ω = 100π Primary voltage is 2295 V, rms and secondary voltage is 143 V, rms Problem 2.2 √ 2Vrms = 167 turns N= ωAc Bpeak Problem 2.3 75 =3 N= turns Problem 2.4 Resistance seen at primary is R1 = (N1 /N2 )2 R2 = 6.25Ω Thus I1 = V1 = 1.6 A R1 and V2 = N2 N1 V1 = 40 V Problem 2.5 The maximum power will be supplied to the load resistor when its impedance, as reflected to the primary of the ideal transformer, equals that of the source (2 kΩ) Thus the transformer turns ratio N to give maximum power must be N= Rs = 6.32 Rload Under these conditions, the source voltage will see a total resistance of Rtot = kΩ and the current will thus equal I = Vs /Rtot = mA Thus, the power delivered to the load will equal Pload = I (N Rload ) = mW 16 Here is the desired MATLAB plot: Problem 2.6 The maximum power will be supplied to the load resistor when its impedance, as reflected to the primary of the ideal transformer, equals that of the source (2 kΩ) Thus the transformer turns ratio N to give maximum power must be N= Rs = 6.32 Rload Under these conditions, the source √ voltage will see a total impedance of Ztot = + j2 kΩ whose magnitude is 2 kΩ The current will thus equal I = √ Vs /|Ztot | = 2 mA Thus, the power delivered to the load will equal Pload = I (N Rload ) = 16 mW Here is the desired MATLAB plot: 17 Problem 2.7 V2 = V1 Xm Xl1 + Xm = 266 V Problem 2.8 part (a): Referred to the secondary Lm,2 = Lm,1 = 150 mH N2 part(b): Referred to the secondary, Xm = ωLm,2 = 56.7 Ω, Xl2 = 84.8 mΩ and Xl1 = 69.3 mΩ Thus, (i) V1 = N Xm Xm + Xl2 V2 = 7960 V and (ii) Isc = V2 V2 = = 1730 A Xsc Xl2 + Xm ||Xl1 Problem 2.9 part (a): I1 = V1 = 3.47 A; Xl1 + Xm V2 = N V1 Xm Xl1 + Xm = 2398 V part (b): Let Xl2 = Xl2 /N and Xsc = Xl1 + Xm ||(Xm + Xl2 ) For Irated = 50 kVA/120 V = 417 A V1 = Irated Xsc = 23.1 V I2 = N Xm Xm + Xl2 Irated = 15.7 A Problem 2.10 IL = Pload = 55.5 A VL and thus IH = IL = 10.6 A; N VH = N VL + jXH IH = 2381 9.6◦ The power factor is cos (9.6◦ ) = 0.986 lagging V 18 Problem 2.11 part (a): part (b): 30 kW jφ e = 93.8 ejφ Iˆload = 230 V A where φ is the power-factor angle Referred to the high voltage side, IˆH = 9.38 ejφ A VˆH = ZH IˆH Thus, (i) for a power factor of 0.85 lagging, VH = 2413 V and (ii) for a power factor of 0.85 leading, VH = 2199 V part (c): 19 Problem 2.12 part (a): part (b): Following methodology of Problem 2.11, (i) for a power factor of 0.85 lagging, VH = 4956 V and (ii) for a power factor of 0.85 leading, VH = 4000 V part (c): Problem 2.13 part (a): Iload = 160 kW/2340 V = 68.4 A at = cos−1 (0.89) = 27.1◦ Vˆt,H = N (VˆL + Zt IL ) which gives VH = 33.7 kV part (b): Vˆsend = N (VˆL + (Zt + Zf )IL ) 20 which gives Vsend = 33.4 kV part (c): ∗ Ssend = Psend + jQsend = Vˆsend Iˆsend = 164 kW − j64.5 kVAR Thus Psend = 164 kW and Qsend = −64.5 kVAR Problem 2.14 Following the methodology of Example 2.6, efficiency = 98.4 percent and regulation = 1.25 percent Problem 2.15 part (a): |Zeq,L | = Vsc,L = 107.8 mΩ Isc,L Req,L = Psc,L = 4.78 mΩ Isc,L Xeq,L = |Zeq,L |2 − Req,L = 107.7 mΩ and thus Zeq,L = 4.8 + j108 mΩ part (b): Req,H = N Req,L = 0.455 Ω Xeq,H = N Xeq,L = 10.24 Ω Zeq,H = 10.3 + j0.46 mΩ part (c): From the open-circuit test, the core-loss resistance and the magnetizing reactance as referred to the low-voltage side can be found: Rc,L = Soc,L = Voc,L Ioc,L = 497 kVA; and thus Voc,L = 311 Ω Poc,L Qoc,L = 2 Soc,L − Poc,L = 45.2 kVAR 21 Xm,L = Voc,L = 141 Ω Qoc,L The equivalent-T circuit for the transformer from the low-voltage side is thus: part (d): We will solve this problem with the load connected to the highvoltage side but referred to the low-voltage side The rated low-voltage current is IL = 50 MVA/8 kV = 6.25 kA Assume the load is at rated voltage Thus the low-voltage terminal voltage is VL = |Vload + Zeq,L IL | = 8.058 kV and thus the regulation is given by (8.053-8)/8 = 0.0072 = 0.72 percent The total loss is approximately equal to the sum of the open-circuit loss and the short-circuit loss (393 kW) Thus the efficiency is given by η= 50.0 Pload = 0.992 = 99.2 percent = Pin 50.39 part (e): We will again solve this problem with the load connected to the high-voltage side but referred to the low-voltage side Now, IˆL = 6.25 25.8◦ kA Assume the load is at rated voltage Thus the low-voltage terminal voltage is VL = |Vload + Zeq,L IˆL | = 7.758 kV and thus the regulation is given by (7.758-8)/8 = -0.0302 = -3.02 percent The efficiency is the same as that found in part (d), η = 99.2 percent Problem 2.16 √ The core length of the second transformer is is times that of the first, its core √ area of the second transformer is twice that of the first, and its volume is 2 times that of the first Since the voltage applied to the second transformer is twice that of the first, the flux densitities will be the same Hence, the core loss will be proportional to the volume and √ Coreloss = 23420 = 9.67 kW 22 The magnetizing inductance is proportional to the area and inversely pro√ portional to the core length√and hence is times larger Thus the no-load magnetizing current will be times larger in the second transformer or √ Ino−load = 4.93 = 6.97 A Problem 2.17 part (a): Rated current at the high-voltage side is 20 kVA/2.4 kV = 8.33 A Thus the total loss will be Ploss = 122 + 257 = 379 W The load power is equal to 0.8 × 20 = 16 kW Thus the efficiency is η= 16 = 0.977 = 97.7 percent 16.379 part (b): First calculate the series impedance (Zeq,H = Req,H + jXeq,H ) of the transformer from the short-circuit test data Req,H = Psc,H = 3.69 Ω Isc,H Ssc,H = Vsc,H Isc,H = 61.3 × 8.33 = 511 kV A Thus Qsc,H = 2 Ssc,H − Psc,H = 442 VAR and hence Xeq,H = Qsc,H = 6.35 Ω Isc,H The regulation will be greatest when the primary and secondary voltages of the transformer are in phase as shown in the following phasor diagram Thus the voltage drop across the transformer will be equal to ∆V = |Iload ||Zeq,H | = 61.2 V and the regulation will equal 61.2 V/2.4 kV = 0.026 = 2.6 percent Problem 2.18 For a power factor of 0.87 leading, the efficiency is 98.4 percent and the regulation will equal -3.48 percent Problem 2.19 part (a): The voltage rating is 2400 V:2640 V part (b): The rated current of the high voltage terminal is equal to that of the 240-V winding, Irated = 30 × 103/240 = 125 A Hence the kVA rating of the transformer is 2640 × 125 = 330 kVA 23 Problem 2.20 part (a): part (b): The rated current of the high voltage terminal is equal to that of the 120-V winding, Irated = 104 /120 = 83.3 A Hence the kVA rating of the transformer is 600 × 83.3 = 50 kVA part (c): The full load loss is equal to that of the transformer in the conventional connection, Ploss = (1 − 0.979) 10 kW = 210 W Hence as an autotransformer operating with a load at 0.85 power factor (Pload = 0.85 × 50 kW = 42.5 kW), the efficiency will be η= 42.5 kW = 0.995 = 99.5 percent 42.71 kW Problem 2.21 part (a): The voltage rating is 78 kV:86 kV The rated current of the high voltage terminal is equal to that of the 8-kV winding, Irated = 50 × 106 /8000 = 6.25 kA Hence the kVA rating of the transformer is 86 kV × 6.25 kA = 537.5 MVA part (b): The loss at rated voltage and current is equal to 393 kW and hence the efficiency will be η= 537.5 MW = 0.9993 = 99.93 percent 538.1 MW Problem 2.22 No numerical result required for this problem Problem 2.23 part (a): 7.97 kV:2.3 kV; 191 A:651 A; 1500 kVA part (b): 13.8 kV:1.33 kV; 109 A:1130 A; 1500 kVA part (c): 7.97 kV:1.33 kV; 191 A:1130 A; 1500 kVA part (d): 13.8 kV:2.3 kV; 109 A:651 A; 1500 kVA Problem 2.24 part (a): (i) 23.9 kV:115 kV, 300 MVA (ii) Zeq = 0.0045 + j0.19 Ω (iii) Zeq = 0.104 + j4.30 Ω 24 part (b): (i) 23.9 kV:66.4 kV, 300 MVA (ii) Zeq = 0.0045 + j0.19 Ω (iii) Zeq = 0.0347 + j1.47 Ω Problem 2.25 Following the methodology of Example 2.8, Vload = 236 V, line-to-line Problem 2.26 The total series impedance is Ztot = Zf + Zt = j11.7 + 0.11 + j2.2 Ω = 0.11 + j13.9 Ω The transformer turns ratio is N = 9.375 The load current, as referred to the transformer high-voltage side will be Iload = N 325 MVA √ 24 kV ejφ = 7.81ejφ kA −1 ◦ where √ φ = − cos 0.93 = −21.6 The line-to-neutral load voltage is Vload = 24 kV part (a): At the transformer high-voltage terminal V = √ |N Vload + Iload Zt | = 231.7 kV, line-to-line part (b): At the sending end V = √ |N Vload + Iload Ztot | = 233.3 kV, line-to-line Problem 2.27 Problem 2.28 First calculate the series impedance (Zeq,H = Req,H + jXeq,H ) of the transformer from the short-circuit test data Zeq,H = 0.48 = j1.18 Ω 25 The total imedance between the load and the sending end of the feeder is Ztot√= Zf + Zeq,H = 0.544 + j2.058 Ω The transformer turns ration is N = 2400:120 = 11.6 part (a): The referred load voltage Vload and current Iload will be in phase and can be assumed to be the phase reference Thus we can write the phasor equation for the sending-end voltage as: Vˆs = Vload + Iload Ztot √ We know that Vs = 2400/sqrt3 = 1386 V and that Iload = 100 kVA/( 32.4 kV) Taking the magnitude of both sides of the above equation gives a quadradic equation in Vload 2 Vload + 2Rtot Iload Vload + |Ztot |2 Iload − Vs2 which can be solved for Vload Vload = −Rtot Iload + Vs2 − (Xtot Iload )2 = 1.338 kV Referred to the low-voltage side, this corresponds to a load voltage of 1.338 kV/N = 116 V, line-to-neutral or 201 V, line-to-line part (b): 2400 = 651 A Feeder current = √ 3Ztot 651 HV winding current = √ = 376 A LV winding current = 651N = 7.52 kA Problem 2.29 part (a): The transformer turns ratio is N = 7970/120 = 66.4 The secondary voltage will thus be V1 Vˆ2 = N jXm R1 + jX1 + jXm = 119.74 0.101◦ part (b): Defining RL = N RL = N kΩ = 4.41 MΩ and Zeq = jXm ||(R2 + RL + jX2 ) = 134.3 + j758.1 kΩ the primary current will equal Iˆ1 = 7970 = 10.3 − 79.87◦ mA R1 + jX1 + Zeq 26 The secondary current will be equal to Iˆ2 = N Iˆ1 jXm R2 + RL + j(Xm + X2 ) = 119.7 0.054◦ mA and thus Vˆ2 = RL Iˆ2 = 119.7 0.054◦ V part (c): Following the methodology of part (b) Vˆ2 = 119.6 0.139◦ V Problem 2.30 This problem can be solved iteratively using MATLAB The minimum reactance is 291 Ω Problem 2.31 part (a): part (b): 27 Problem 2.32 part (a): The transformer turns ratio N = 200/5 = 40 For I1 = 200 A I2 = I1 N jXm R2 + j(Xm + X2 ) = 4.987 0.024◦ part (b): Defining RL = N 250µΩ = 0.4 Ω I2 = I1 N jXm R2 + RL + j(Xm + X2 ) = 4.987 0.210◦ Problem 2.33 part (a): part (b): Problem 2.34 Zbase,L = Zbase,H = Vbase,L Pbase Vbase,H Pbase = 1.80 Ω = 245 Ω 28 Thus R1 = 0.0095Zbase,L = 17.1 mΩ; X1 = 0.063Zbase,L = 113 mΩ Xm = 148Zbase,L = 266 Ω R2 = 0.0095Zbase,H = 2.33 Ω; X2 = 0.063Zbase,H = 15.4 Ω Problem 2.35 part (a): (i) Zbase,L = (7.97 × 103 )2 = 0.940 Ω; 75 × 103 (ii) Zbase,H = (7970)2 = 847 Ω; 75 × 103 XL = 0.12Zbase,L = 0.113 Ω XH = 0.12Zbase,H = 102 Ω part (b): (i) 797 V:13.8 kV, 225 kVA (ii) Xpu = 0.12 (iii) XH = 102 Ω (iv) XL = 0.339 Ω part (c): (i) 460 V:13.8 kV, 225 kVA (ii) Xpu = 0.12 (iii) XH = 102 Ω (iv) XL = 0.113 Ω Problem 2.36 part (a): In each case, Ipu = 1/0.12 = 8.33 pu √ √ (i) Ibase,L = Pbase /( Vbase,L ) = 225 kVA/( 797 V) = 163 A IL = Ipu Ibase,L √ = 1359 A √ (ii) Ibase,H = Pbase /( Vbase,H ) = 225 kVA/( 13.8 kV) = 9.4 A IH = Ipu Ibase,H = 78.4 A part (b): In each case, Ipu = 1/0.12 = 8.33 pu √ √ (i) Ibase,L = Pbase /( Vbase,L ) = 225 kVA/( 460 V) = 282 A IL = Ipu Ibase,L √ = 2353 A √ (ii) Ibase,H = Pbase /( Vbase,H ) = 225 kVA/( 13.8 kV) = 9.4 A IH = Ipu Ibase,H = 78.4 A 29 Problem 2.37 part (a): On the transformer base Xgen = Pbase,t Pbase,g 1.57 = 800 MVA 850 MVA 1.57 = 1.27 pu part (b): On the transformer base, the power supplied to the system is Pout = 700/850 = 0.824 pu and the total power is Sout = Pout /pf = 0.825/0.95 = 0.868 pu Thus, the per unit current is Iˆ = 0.868 φ, where φ = − cos−1 0.95 = −18.2◦ (i) The generator terminal voltage is thus ˆ t = 1.03 3.94◦ pu = 26.8 3.94◦ kV Vˆt = 1.0 + IZ and the generator internal voltage is ˆ t + Zgen ) = 2.07 44.3◦ pu = 53.7 44.3◦ kV Vˆgen = 1.0 + I(Z (ii) The total output of the generator is given by Sgen = Vˆt Iˆ∗ = 0.8262 + 0.3361 Thus, the generator output power is Pgen = 0.8262 × 850 = 702.2 MW The correspoinding power factor is Pgen /|Sgen | = 0.926 lagging ... I1 = 20 0 A I2 = I1 N jXm R2 + j(Xm + X2 ) = 4.987 0. 024 ◦ part (b): Defining RL = N 25 0µΩ = 0.4 Ω I2 = I1 N jXm R2 + RL + j(Xm + X2 ) = 4.987 0 .21 0◦ Problem 2. 33 part (a): part (b): Problem 2. 34... V Problem 2. 30 This problem can be solved iteratively using MATLAB The minimum reactance is 29 1 Ω Problem 2. 31 part (a): part (b): 27 Problem 2. 32 part (a): The transformer turns ratio N = 20 0/5... N Xm Xm + Xl2 V2 = 7960 V and (ii) Isc = V2 V2 = = 1730 A Xsc Xl2 + Xm ||Xl1 Problem 2. 9 part (a): I1 = V1 = 3.47 A; Xl1 + Xm V2 = N V1 Xm Xl1 + Xm = 23 98 V part (b): Let Xl2 = Xl2 /N and Xsc